# Let $f$ be a real valued function on the interval $(0,1)$. Prove that the set of points of discontinuity of the _first kind_ is at most countable.

Let $f$ be a real valued function on the interval $(0,1)$. Prove that the set of points of discontinuity of the first kind is at most countable.

Definition $x_0$ is a point of discontinuity of the first kind of $f$ if $$\lim_{x \to x_0^+} f(x) > \lim_{x \to x_0^-} f(x)$$ and both of the above limits exist.

There is a straight forward way to show this fact for a monotone function (using the following Hint), but I do not know how to proceed for the general case.

Hint For each $n\in \mathbb{N}$ show that the set of points $x$ for which $$|\lim_{t \to x^+} f(t) - \lim_{t \to x^-} f(t)| > \frac{1}{n}$$ is at most countable.

Call $x_0$ a $\frac1n$-discontinuity of the first kind if $$|\lim_{x \to x_0^+} f(x) - \lim_{x \to x_0^-} f(x)| > \frac{1}{n}$$ Since the left and right limits exist, there exists $\epsilon>0$ such that $|f(y)-\lim_{x \to x_0^-} f(x)| < \frac1{2n}$ if $y \in (x_0-\epsilon,x_0)$, and $|f(y)-\lim_{x \to x_0^+} f(x)| < \frac1{2n}$ if $y \in (x_0,x_0+\epsilon)$. Therefore any $y \in (x_0-\epsilon,x_0+\epsilon)$ cannot be a $\frac1n$-discontinuity of the first kind.
• How can you assume there is such $y$ ? Are you assuming piecewise continuity ? – the8thone Jan 5 '14 at 3:05